Homology with twisted coefficients of the circle $\DeclareMathOperator{\Aut}{Aut}
\newcommand{\p}{\pi_1X} 
\newcommand{\Z}{\mathbb{Z}} 
\newcommand{\XX}{\tilde{X} } $
Let $X$ be a path connected and locally path connected topological space and $\rho:\p\to \Aut(A)$ a representation of the fundamental group  into an abelian group $A$. We define 
$$H_k(X;A_\rho),$$ the homology of $X$ with coefficients in $A$ twisted by $\rho$, by taking the homology of the chain complex
$$ C_{\bullet}(\tilde{X}) \;\underset{\Z[\p]}\otimes\; A $$
where:


*

*$\Z[\p]$ is the group ring of the fundamental group

*$\tilde{X}$ is the universal cover of $X$ 

*$C_{\bullet}(\XX)$ is the singular chain complex of $\tilde{X}$, endowed with the natural right $\Z[\p]$-module structure given by the monodromy action. That is: any $g\in \p$ acts naturally on $\XX$ as a deck transformation $g:\XX\to\XX$. Thus given a $k$-simplex $\sigma: \Delta^k\to \XX$ we define
$$ \sigma \cdot g := g\circ\sigma : \Delta^k\to \XX \to \XX $$

*The $\mathbb{Z}[\p]$-module structure on $A$ is given by the representation $\rho$.
I am trying to compute $H_1(X;A_\rho)$ when $X=S^1$ and $A=\Z_3$, with twist given by the nontrivial representation $\rho:\mathbb{Z}\to \Aut(\Z_3)$, i.e. the map given by
$$ \rho(1) = \begin{cases} 0 \mapsto 0 \\ 1 \mapsto 2 \\ 2 \mapsto 1 \end{cases} $$

I have the geometrical picture of what's going on in my mind, but I'm having troubles with the algebra. 
What I did so far:
The CW structure of $S^1$ can be pulled back to $\XX \cong \mathbb{R}$, and we obtain the cellular chain complex (as $\Z[\p]$-module)
$$ 0\to \Z[\p] \to \Z[\p] \to 0, $$
and since $\Z[\p] \cong \Z[t, t^{-1}]$ the above is
$$ 0\to \Z[t, t^{-1}] \to \Z[t, t^{-1}] \to 0. $$
The boundary map is given by multiplication by $t-1$.
I need help:
I can't figure out what happens to the boundary map when we apply to the above chain complex the functor $$-\;\underset{\Z[\p]}\otimes\;\Z_3$$
Can you help me with that, and thus give me the final result for this twisted homology groups? 
Moreover, can this be generalized easily to an arbitrary abelian group $A$ and any representation $\rho:\p\to \Aut(A)$?
 A: $\DeclareMathOperator{\Aut}{Aut}
\newcommand{\p}{\pi_1X} 
\newcommand{\Z}{\mathbb{Z}} 
\newcommand{\XX}{\tilde{X} }$Ok, I finally figured it out! So, the cellular chain complex of $\XX$ is given by
$$ 0\longrightarrow \Z[t, t^{-1}] \overset{\delta}\longrightarrow \Z[t, t^{-1}] \longrightarrow 0, $$
where the boundary map $\delta$ is just multiplication by $t-1$, for geometrical reasons.
The new boundary map $D:\Z_3\to\Z_3$ which we obtain afte tensoring with $\Z_3$ over $\Z[t,t^{-1}]$ can be computed by looking at the sequence of maps
$$
\Z_3
\overset{\cong}\longrightarrow
\Z[t,t^{-1}]\underset{\Z[t,t^{-1}]}\otimes\Z_3 \;
\overset{\delta\otimes id}\longrightarrow 
\Z[t,t^{-1}]\underset{\Z[t,t^{-1}]}\otimes\Z_3 
\overset{\cong}\longrightarrow
\Z_3,
$$
where the first map is $a\mapsto 1\otimes a$ and last map is $1\otimes a \mapsto a $. Of course the point is that we need reduce to the form $1\otimes a$ before applying the last map. Therefore we get
$$ 
a \mapsto
1\otimes a 
\mapsto (t-1)\otimes a = 1 \otimes (ta - a) 
\mapsto ta - a. 
$$
Hence by a direct computation $D$ turns out to be the identity of $\Z_3$:
$$ D(0) = 0, \quad D(1)=t\cdot1 - 1 = 2-1 = 1, \quad D(2)=t\cdot 2 - 2 = 1-2 \equiv 2 \mod 3. $$
Therefore we conclude that the homology groups of $S^1$ with coefficients in $\Z_3$ twisted by the nontrivial $\rho$ are trivial:
$$ H_0(S^1\Z_3)_\rho \cong H_1(S^1\Z_3)_\rho \cong 0   $$
